Why not just use Bell state entanglement in the CHSH game?

Question

In the CHSH game, both Alice and Bob receive random bits $x$ and $y$ from a referee Charlie. Based on the bit values and a strategy discussed between Alice and Bob beforehand they will respond with bit values $a$ and $b$. During the game, Alice and Bob cannot communicate. The goal of the game is to produce matching bits $a$ and $b$.

The best possible classical strategy is for both Alice and Bob to always respond with a 0, which leads to a $3/4$ success probability.

In the quantum case, Alice and Bob share an entangle qubit in the state $\psi = 1/\sqrt{2}(|0_A0_B\rangle + |1_A1_B\rangle)$. When Alice receives a bit $x = 0$ she measures in the $|0\rangle, |1\rangle$ basis and if she gets $x = 1$ she measures in the $|+\rangle, |-\rangle$ basis. Correspondingly, if Bob receives $y = 0$ he measure in $|a_0\rangle, |a_1\rangle$, where $a_0 = \cos(\pi/8)|0\rangle + \sin(\pi/8)|1\rangle$. If he receives $y = 1$ he measure in $|b_0\rangle, |b_1\rangle$, where $b_0 = -\sin(\pi/8)|0\rangle + \cos(\pi/8)|1\rangle$

I understand the math behind the CHSH game and how the success probability increases to $\cos(\pi/8)^2$. However, Alice and Bob have a Bell pair in the state $\psi = 1/\sqrt{2}(|00\rangle + |11\rangle)$. Doesn't this mean that if Alice measures $|0\rangle$ so will Bob, and similar for $|1\rangle$. I believe techniques like quantum teleportation are based on this principle.

Why don't they just agree on a strategy to report back to the referee whatever they have measured? This would lead to a $100\%$ success rate. I must be missing something.

Answer 1

"The goal of the game is to produce matching bits $a$ and $b$."

This is only true when the inputs are not both equal to 1. When the inputs are both equal to 1 then they are required to produce opposite bits to win. This is why the classical strategy only wins 3/4 of the time (1/4 of the time they both get input 1 and the strategy fails).

The quantum strategy you propose in your question will also fail for this reason and only achieve a winning probability of 3/4. If Alice and Bob always both measure in the computational basis and announce their outcomes they'll always announce the same outcome (never opposite).